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A family of positively many $s$-cliques forces a logarithmic balanced graph blow-up, which can also contain a matching of that many members of the family. To induct on $s$, prune faces with few extensions, apply induction to the remaining $(s-1)$-faces, and use a matching of those faces as one side of a bipartite incidence graph. The common neighbourhood from bipartite density estimate selects logarithmically many disjoint faces with polynomially many common extension vertices.

A fixed positive excess over the Turan theorem threshold forces logarithmic balanced complete multipartite graphs. There is an elementary induction using common neighbourhood from bipartite density: first obtain a logarithmic $r$-partite subgraph from the induction hypothesis, and select disjoint transversal $r$-cliques in it. Each such clique has at least $r\epsilon n$ common neighbours, by the union bound on non-neighbour sets. Apply the bipartite common-neighbour count to these cliques as left-side objects. Their selected union and a large common neighbourhood form the extra part.

Write $K_s(t)$ for a balanced complete multipartite blow-up with $s$ classes of size $t$. In the nonvacuous range $0<\epsilon <1/r$, the answer isWe prove both bounds, including the quantitative lower bound.

First, suppose a bipartite graph with parts $A,B$, of sizes $m,N$, has at least $pmN$ edges. For any integer $s\le pm/2$, counting pairs consisting of an $s$-subset of $A$ and a common neighbour in $B$ givesThe inequality follows from discrete convexity: $\left(\genfrac{}{}{0ex}{}{x+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{x}{s}\right)=\left(\genfrac{}{}{0ex}{}{x}{s-1}\right)$ is nondecreasing, so balancing integer degrees of vertices minimizes their sum. Consequently some $S$ hasThe product estimate uses $\lfloor pm\rfloor -j\ge pm-s$ for $0\le j<s$. This is a common neighbourhood from bipartite density bound.

We next prove by induction on $q\ge 1$ that, for every fixed $\alpha >0$, a graph $J$ of order $N$ with minimum degree at least $(1-1/q+\alpha )N$ contains $K_{q+1}(\lfloor c\log N\rfloor )$, for some $c=c(q,\alpha )>0$ and all sufficiently large $N$. Impossible minimum-vertex degree parameters need no argument. For $q=1$, take two copies of $V(J)$ with adjacency given by $J$. The preceding count with $p=\alpha$ and $s=\lfloor c\log N\rfloor$, where $c\log (2/\alpha )<1/2$, gives at least $\sqrt{N}$ common neighbours. The selected vertices and their common neighbours are disjoint in $J$, since a vertex is not its own neighbour. Selecting $s$ common neighbours gives $K_2(s)$.

For $q\ge 2$, the minimum degree hypothesis implies the induction hypothesis for $q-1$, for example with excess $1/[q(q-1)]$. Thus $J$ contains $K_q(m)$ with $m=\lfloor a\log N\rfloor$ for a positive constant $a$. Pair the vertices in its classes into $m$ disjoint transversal cliques $C_1,\dots ,C_m$, each of order $q$. By the union bound, every $C_i$ has at leastcommon neighbours. Form an auxiliary bipartite graph between these $m$ cliques and the $N$ vertices, using common-neighbour incidence, and put $p=q\alpha$. Choose $c>0$ small enough that $c<pa/3$ and $c\log (2/p)<1/2$. The count above, with $s=\lfloor c\log N\rfloor$, gives $s$ cliques with at least $\sqrt{N}$ simultaneous common neighbours. Their vertices form $q$ classes of size $s$, and $s$ of those common neighbours form the last class. None of the common neighbours belongs to a selected clique, because that would require a loop. This proves the logarithmic clique blow-up from minimum degree assertion.

To pass from an edge count to minimum degree, put $b=1-1/r+\epsilon$ and $a=b-\epsilon /2$. Repeatedly remove a vertex whose current degree of a vertex is below $a$ times the current order. If the process reaches order $m$, its deleted edges and remaining edges giveThus $(1-a)m^2\ge (\epsilon /2)n^2-O(n)$. In particular, for a fixed sufficiently small $\gamma >0$, the process cannot reach $m=\lfloor \gamma n\rfloor$: the displayed inequality would fail. It stops earlier with a subgraph of order $N\ge \gamma n$ and minimum degree at least $(1-1/r+\epsilon /2)N$. The induction just proved supplies $K_{r+1}(\lfloor c\log N\rfloor )$, hence a lower bound $c^{\prime }\log n$ on $t(r,\epsilon ,n)$. In particular, the guaranteed part size tends to infinity.

For the upper bound, choose a fixed $p$ with $b<p<1$ and take a binomial random graph $G(n,p)$. Its edge count has expectation $p\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and variance $p(1-p)\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. The Chebyshev inequality shows that $e(G)\ge b\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ with probability tending to one. The expected number of labeled copies of $K_{r+1}(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/[r\log (1/p)]$, this tends to zero. The Markov inequality therefore shows that with positive probability both the required edge density holds and there is no such copy. This proves $t(r,\epsilon ,n)=O(\log n)$ and completes the random obstruction to larger clique blow-ups argument.

There is a necessary qualification to the density parameter. If $\epsilon =1/r$, the only eligible graph is the complete graph, and $t(r,1/r,n)=\lfloor n/(r+1)\rfloor$. If $\epsilon >1/r$ and $n\ge 2$, there are no eligible simple graphs; the universal assertion is vacuous for every $t$, so the printed maximum has no finite value in that range.

Write $K_j(t)$ for a balanced complete multipartite blow-up, with $j$ parts each containing $t$ vertices. Its occurrence is a subgraph condition, so additional internal edges in the host are harmless. The coefficient multiplying $\log n$ is essential: the PDF has $t=\lfloor d\log n\rfloor$, although it is lost in the converted TeX.

First prove the common neighbourhood from bipartite density estimate. Suppose a bipartite graph has parts of sizes $m,N$ and at least $\beta mN$ edges. For an integer $s\le \beta m/2$, count incidences between $s$-subsets of the first part and their common neighbours in the second. The discrete convexity of $k\mapsto \left(\genfrac{}{}{0ex}{}{k}{s}\right)$, with value zero for $k<s$, givesThe convexity assertion follows directly from $\left(\genfrac{}{}{0ex}{}{k+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{k}{s}\right)=\left(\genfrac{}{}{0ex}{}{k}{s-1}\right)$, which is increasing in $k$; smoothing two unequal degrees cannot increase the sum. In the binomial ratio every factor is at least $\beta /2$, because $\lfloor \beta m\rfloor -s+1\ge \beta m-s\ge \beta m/2$. Consequently some $S$ satisfies

We prove the logarithmic clique blow-up from minimum degree by induction on $r$. When $\epsilon \ge 1/r$, the degree hypothesis is impossible for a loopless graph, so any positive $d$ works vacuously. Assume $0<\epsilon <1/r$. For $r=1$, take two copies of the vertex set and join the copies of adjacent vertices. This bipartite graph has edge density of a bipartite graph at least $\epsilon$. ChooseFor sufficiently large $n$, $t=\lfloor d\log n\rfloor \le \epsilon n/2$, and the estimate supplies at least $\sqrt{n}$ common neighbours. Choose $t$ of them. The two selected sets are disjoint in the original graph, since a selected vertex cannot be adjacent to itself. They give $K_2(t)$.

For $r\ge 2$, put ${\epsilon }^{\prime }=\epsilon +1/(r-1)-1/r$. The same minimum degree of a graph exceeds $(1-1/(r-1)+{\epsilon }^{\prime })n$. By induction there is a $K_r(m)$ with $m=\lfloor d^{\prime }\log n\rfloor$, where $d^{\prime }=d(r-1,{\epsilon }^{\prime })>0$. Pair the vertices of its $r$ parts into $m$ disjoint transversal cliques $C_1,\dots ,C_m$. Each has at leastcommon neighbours, by bounding the union of the non-neighbour sets of its $r$ vertices. Set $\beta =\min \{r\epsilon ,1/2\}$. Form the bipartite graph whose first part is these $m$ cliques, with $C_i$ joined to a vertex when that vertex is a common neighbour of all of $C_i$. It has density at least $\beta$. ChooseFor large $n$, $m\ge d^{\prime }\log n/2$ and $t=\lfloor d\log n\rfloor \le \beta m/2$. The estimate gives $t$ selected cliques and at least $\sqrt{n}\ge t$ common vertices. The selected cliques supply $t$ vertices in each of the original $r$ parts, and their common vertices supply the final part. No common vertex belongs to a selected clique, again because there are no loops. All required cross-part edges are present. Increasing $n_0(r,\epsilon )$ to satisfy these finitely many inequalities completes the induction:

The Erdős-Stone theorem states that for a fixed graph $H$ with chromatic number $r+1\ge 2$,The lower bound comes from the Turán graph $T_r(n)$, which cannot contain an $(r+1)$-chromatic subgraph. To obtain the upper bound from our minimum-degree result, suppose $e(G)\ge (a+\eta )n^2/2$, where $a=1-1/r$ and $0<\eta <1-a$. Repeatedly delete a vertex of degree less than $b$ times the current order, where $b=a+\eta /2$, until no such vertex remains. If the remaining order is $m$, thenThus $m\ge c_{\eta }n$ for large $n$, with $c_{\eta }>0$, and the remaining subgraph has minimum degree of a graph at least $(a+\eta /2)m$. It contains $K_{r+1}(\lfloor d\log m\rfloor )$. A proper graph colouring of $H$ places its vertices in the $r+1$ parts once their size exceeds $|V(H)|$. This contradicts $H$-freeness. Adjusting a fixed positive density margin absorbs the difference between $n^2/2$ and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$, proving the upper bound.

The six-vertex forbidden graph is the octahedral graph $K_{2,2,2}$: the deleted matching in a graph gives its three independent pairs. Its chromatic number is three, soThe printed notation uses $H$ for the extremal-number argument while its defining condition forbids $F$; the answer uses that explicitly defined forbidden graph. For the strict finite inequality, take $T_2(n)$ and add one edge inside its larger part, which has at least two vertices for $n\ge 3$. Every triangle in a graph then uses this added edge. However, $K_{2,2,2}$ contains two vertex-disjoint triangles in a graph, obtained by taking one vertex from each pair and then the other three. It therefore cannot occur in the constructed graph. Hence

Write $K_s(t)$ for the balanced graph blow-up of a complete graph, with $s$ classes of size $t$; containment here is as a subgraph, not necessarily an induced subgraph. The logarithmic form of the Erdős-Stone theorem is: for fixed $r\ge 1$ and $0<ϵ<1/r$, there are $d=d(r,ϵ)>0$ and $n_0$ such thatThus the guaranteed balanced part size is at least $\lfloor d\log n\rfloor$, with $d$ independent of $n$. We prove the logarithmic Erdős-Stone theorem by a density-to-cliques step and a constructive dense clique family blow-up lemma.

First choose a fixed $m$ large enough that $t_r(m)/\left(\genfrac{}{}{0ex}{}{m}{2}\right)\le 1-1/r+ϵ/2$, where $t_r(m)$ is the edge count of the Turán graph. A uniformly chosen $m$-vertex subset has expected edge count at least $(1-1/r+ϵ)\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. If its induced graph contains no $K_{r+1}$, the Turan theorem bounds that count by $t_r(m)$. Since the count is always at most $\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the probability that the subset contains an $(r+1)$-clique is at least $ϵ/2$. Double counting the incidences between such subsets and their cliques gives, for $s=r+1$,for a fixed $\eta >0$ and all sufficiently large $n$. Here $(x{)}_s=x(x-1)\cdots (x-s+1)$. This is clique supersaturation by sampling.

We now prove the needed dense clique family blow-up lemma, with the following stronger induction invariant. Given any family $\mathcal{M}$ of at least $\eta n^s$ distinct $s$-cliques, there is a complete $s$-partite subgraph with each class of size $\lfloor a_s(\eta )\log n\rfloor$, containing that many pairwise vertex-disjoint transversal members of $\mathcal{M}$, each with one vertex in every class. Extra edges within its classes can be ignored. We may suppose $0<\eta <1/2$. For $s=1$, any $\eta n$ singletons suffice, and we may take $a_1(\eta )=1$ once $n$ is large.

For $s\ge 2$, put $\theta =\eta /2$. Repeatedly delete every member of the current family containing an $(s-1)$-clique whose number of extensions is at most $\theta n$. Each such face is processed at most once, and there are at most $n^{s-1}$ possible faces. At most $\theta n^s$ members are deleted. The remaining family $\mathcal{L}$ has size at least $(\eta /2)n^s$, and every face that remains has more than $\theta n$ extensions. Its family of $(s-1)$-faces has size at least $s|\mathcal{L}|/n\ge (\eta /2)n^{s-1}$, since each face is in at most $n$ members.

Apply the induction hypothesis to these faces. It gives a complete $(s-1)$-partite subgraph with $m=\lfloor a_{s-1}(\eta /2)\log n\rfloor$ vertices in each class and a matching $R_1,\dots ,R_m$ of transversal $(s-1)$-faces from the family. Form a bipartite graph whose left vertices are the $R_i$ and whose right vertices are the original $n$ vertices; join $R_i$ to $v$ when $R_i\cup \{v\}\in \mathcal{L}$. Every left degree exceeds $\theta n$.

The following common neighbourhood from bipartite density estimate is elementary. In a bipartite graph with class sizes $m,n$ and at least $\theta mn$ edges, averaging over left subsets $S$ of size $u$ and using convexity of the integer sequence $j\mapsto \left(\genfrac{}{}{0ex}{}{j}{u}\right)$ giveswhenever $1\le u\le \theta m/2$. The last inequality follows by comparing the $u$ factors in the two binomial coefficients; the integer convexity follows from the nondecreasing first differences $\left(\genfrac{}{}{0ex}{}{j}{u-1}\right)$.

ChooseFor large $n$, $m\ge \frac{1}{2}{a}_{s-1}(\eta /2)\log n$, so $u\le \theta m/4$. The estimate supplies $u$ selected faces with a common extension set of size at least $n(\theta /2{)}^u\ge \sqrt{n}$. This extension set is disjoint from the selected faces: a vertex in any one of them cannot extend that face. Their union has $u$ vertices in each of the old $s-1$ classes, with all required cross edges; choose $u$ distinct common extension vertices as a new class. Attaching one different extension vertex to each selected face also gives $u$ disjoint members of $\mathcal{L}$. This completes the induction, with positive constants independent of $n$. Apply it to the clique family above and take $d=a_{r+1}(\eta )$.

The logarithmic order cannot be increased in a uniform forcing result. Put $\rho =1-1/r+ϵ<1$, choose $p$ strictly between $\rho$ and one, and take a binomial random graph. Its edge density exceeds $\rho$ with probability tending to one, by the variance estimate for a sum of independent edge indicators. With $s=r+1$, the expected number of ordered embeddings of $K_s(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/((s-1)\log (1/p))$, the logarithm of this bound is negative of order $(\log n{)}^2$. Markov's inequality therefore makes the probability of any such copy tend to zero. Both events hold simultaneously for some graphs of every sufficiently large order. There are graphs meeting the density hypothesis whose largest balanced blow-up has part size $O(\log n)$. This upper bound concerns what density can force; it is not an upper bound for every dense graph, since a complete graph has much larger blow-ups.

Finally, $H_k$ is the square of a cycle, for $k\ge 3$. Any three consecutive vertices form a triangle in a graph. In a proper three-colouring, once the first three colours are fixed, each subsequent vertex must repeat the colour three positions earlier. Closing the cycle is possible precisely when $3\mid k$. Conversely, repeating the three colours gives such a colouring whenever $3\mid k$. In that case $H_k$ embeds in $K_3(k/3)$, and the density $0.51\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ exceeds the bipartite Turan theorem threshold by a fixed amount, so the theorem forces $H_k$ eventually. If $3\nmid k$, the balanced Turán graphs $T_3(n)$ have density at least $2/3$ and contain no graph of chromatic number greater than three. They give a counterexample sequence. Exactly the cycle lengths $k\ge 3$ divisible by three have the required property.